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en/docs/chapter_greedy/exercises.md

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+---
+comments: true
+---
+
+<!-- Generated by utils/exercises/publish_exercises.py; do not edit directly. -->
+
+# 15.6 &nbsp; Exercises
+
+## 15.6.1 &nbsp; Concept Review
+
+### 1. &nbsp; Is Choosing the Largest Coin Always Best?
+
+The coin denominations are `[1, 7, 10]`, and the target amount is 14.
+
+<!-- numbered-subquestions -->
+
+1. Follow the rule "always choose the largest denomination that does not exceed the remaining amount," and write the coins selected.
+2. Is there a solution using fewer coins? If so, give one; otherwise, explain why not.
+3. Does this example show that the greedy strategy is correct for every set of coin denominations?
+
+??? success "Answer"
+
+    1. The greedy strategy selects `10 + 1 + 1 + 1 + 1`, using 5 coins in total.
+
+    2. There is a solution using fewer coins: `7 + 7`, which uses only 2 coins.
+
+    3. No. This counterexample shows that, for arbitrary coin denominations, repeatedly choosing the largest currently available denomination does not necessarily minimize the number of coins.
+        The largest immediate choice may prevent a better combination later.
+
+### 2. &nbsp; Which Item Should Go into the Knapsack First?
+
+A knapsack with a capacity of 4 kilograms can hold the following items. A fraction of an item may be taken,
+and the value obtained is proportional to its weight:
+
+- Item A: weight 4 kilograms, value 20.
+- Item B: weight 3 kilograms, value 18.
+
+<!-- numbered-subquestions -->
+
+1. What is the value per kilogram of each item? Which item should be placed in the knapsack first?
+2. Fill the knapsack using the greedy strategy for the fractional knapsack problem. What is the final value?
+3. When items can be divided and the knapsack limits total weight, should items be compared by total value or by value per kilogram? Why?
+
+??? success "Answer"
+
+    1. A has value `20 ÷ 4 = 5` per kilogram, while B has value `18 ÷ 3 = 6` per kilogram,
+        so B, with the higher value per unit weight, should be placed in the knapsack first.
+
+    2. First take all of B, using 3 kilograms of capacity and gaining a value of 18. With 1 kilogram of capacity remaining,
+        take 1 kilogram of A, gaining a value of 5. The final value is `18 + 5 = 23`.
+
+    3. The knapsack limits total weight, and items can be divided, so they should be compared by value per unit weight.
+        Although A has a higher total value, its value per kilogram is lower than B's. Filling the knapsack with A first would yield only a value of 20.
+
+### 3. &nbsp; Which Pointer Should Move Next?
+
+The partition heights are `[1, 8, 6, 2, 5]`. Use two pointers, one at each end, to find the maximum capacity.
+The capacity equals "the height of the shorter partition × the distance between the partitions' indices."
+
+<!-- numbered-subquestions -->
+
+1. Initially, the left pointer is at index 0 and the right pointer is at index 4. What is the current capacity? Which pointer should move next?
+2. After making the move chosen in Question 1, at which indices are the two pointers? What is the capacity now? Which pointer should move next?
+3. For the current pair of partitions, you could move either the pointer at the shorter partition or the pointer at the taller partition. Which move could still produce a greater capacity, and why?
+
+??? success "Answer"
+
+    1. The current capacity is `min(1, 5) × (4 - 0) = 4`. The left partition is shorter, so move the left pointer.
+
+    2. After the left pointer moves, the pointers are at indices 1 and 4. The current capacity is
+        `min(8, 5) × (4 - 1) = 15`. The right partition is shorter, so move the right pointer next.
+
+    3. Moving the pointer at the shorter partition is the only move that could still produce a greater capacity. If the taller partition's pointer moves, the distance certainly decreases while the height remains limited by the unmoved shorter partition,
+        so the capacity can only stay the same or decrease. Only by moving the shorter partition can a taller partition possibly be found.
+
+## 15.6.2 &nbsp; Programming Exercises
+
+### 1. &nbsp; Fractional Knapsack
+
+You are given equal-length arrays `wgt` and `val`, where `wgt[i] > 0` and `val[i] >= 0`. The knapsack has capacity `cap >= 0`.
+There is only one of each item, but any fraction of an item may be placed in the knapsack.
+The value obtained is proportional to the fraction of the item's total weight that is included. Use a greedy algorithm
+and return the maximum total value the knapsack can hold as a real number.
+
+??? tip "Hints"
+
+    1. First calculate each item's value per unit weight as val[i] / wgt[i], keeping the fractional part of the division
+    2. Place items with higher value per unit weight into the knapsack first
+    3. If the remaining capacity is less than the current item's weight, take exactly the fraction that fills the knapsack and stop

en/docs/chapter_greedy/index.md

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 - [15.3   Maximum Capacity Problem](max_capacity_problem.md)
 - [15.4   Maximum Product Cutting Problem](max_product_cutting_problem.md)
 - [15.5   Summary](summary.md)
+- [15.6   Exercises](exercises.md)